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EE 205 Dr. A. Zidouri Electric Circuits II The Transformer Lecture #23 -1- EE 205 Dr. A. Zidouri The material to be covered in this lecture is as follows: o Introduction to Transformers o Analysis of a Linear Transformer Circuit o Reflected Impedance After finishing this lecture you should be able to: ¾ ¾ ¾ ¾ Construct the frequency domain equivalent circuit of a transformer Analyze a linear transformer circuit Calculate the reflected impedance into the primary Calculate the scaling factor for the reflected impedance -2- EE 205 Dr. A. Zidouri Introduction ¾ A Transformer is a magnetic device that takes advantage of the phenomenon of mutual inductance. ¾ A Transformer is generally a four-terminal device comprising two (or more) magnetically coupled coils. ¾ A Transformer may also be regarded as one whose flux is proportional to the currents in its windings. Consider the following general circuit model for a transformer used to connect a load to a source. jωMI2 jωMI1 ZS VS a I1 R1 jωM jωL1 b R2 c I2 jωL2 d Fig.33-1 Frequency-domain circuit model for a Transformer -3- EE 205 Dr. A. Zidouri Terminology ¾ The transformer winding connected to the source is the primary winding. ¾ The transformer winding connected to the load is the secondary winding. Where: R1 → The resistance of the primary winding R2 → The resistance of the secondary winding L1 → The self inductance of the primary winding L2 → The self inductance of the secondary winding M → The mutual inductance VS → The internal Voltage of the sinusoidal source ZS → The internal impedance of the sinusoidal source ZL → The Load impedance. -4- EE 205 Dr. A. Zidouri Analysis of a linear transformer circuit The analysis of the circuit consists of finding I1 and I2 as a function of circuit parameters: R1, R2, L1, L2, M, VS, ZS, ZL and ω. Mesh 1 VS = ( ZS +R1 + jωL1 ) I1 - jωMI2 (33-1) Mesh 2 0 = -jωMI1 + ( ZL +R2 + jωL2 ) I2 (33-2) For simplification, let Z11 = R1 + jωL1 + ZS Z22 = R2 + jωL 2 + ZL Self impedance of the primary (33-3) Self impedance of the secondary (33-4) Thus, VS = Z11I1 - jωMI2 0 = -jωMI1 + Z 22I2 From (33-6) yields: I2 = (33-5) (33-6) jωM I1 Z 22 (33-7) Substitute in (33-5) yields -5- EE 205 Dr. A. Zidouri ⎛ Z11Z 22 + ω2M2 ω2M2 ⎞ VS = I1 = ⎜ Z11 + ⎟ I1 Z 22 Z 22 ⎠ ⎝ (33-8) Therefore Z 22 VS Z11Z 22 + ω2M2 jωM jωM I2 = VS = I1 2 2 Z11Z 22 + ω M Z 22 I1 = Primary Current (33-9) Secondary Current (33-10) The impedance at the terminals of the source is: Z int - Z S so, ω2M2 ω2M2 Z ab = Z11 + − Z S = R1 + jωL1 + Z 22 R2 + jωL 2 + ZL Impedance looking into the primary (33-11) Note: The impedance Zab is independent of the magnetic polarity of the transformer. -6- EE 205 Dr. A. Zidouri Reflected Impedance ω2M2 The quantity Zr = R2 + jωL 2 + ZL is called the Reflected Impedance (33-12) This is due solely to the existence of mutual inductance. To consider reflected impedance in more detail, we express the load impedance in rectangular form: ZL = RL + jXL (33-13) where XL is positive for inductive load and negative for capacitive load. Zr in rectangular form will be: ω2M2 ⎡⎣(RL +R 2 ) - j ( ωL 2 + XL ) ⎤⎦ Zr = 2 2 (RL +R2 ) + ( ωL2 + XL ) Zr = ω2M2 Z 22 2 ⎡⎣(RL +R2 ) - j ( ωL 2 + XL ) ⎤⎦ = (33-14) ω2M2 Z 22 2 Z∗22 (33-15) Therefore the Linear Transformer reflects the conjugate of the self-impedance of the secondary circuit ( Z 22 ) ∗ into the primary circuit. The coefficient ω2M2 Z 22 2 is called the scaling factor of the reflected impedance. -7- EE 205 Dr. A. Zidouri Example 33-1 Calculate the reflected impedance, the scaling factor, the primary and secondary currents for the Circuit of Fig.33-2: Parameters of the transformer R 2 = 40 Ω, X1 = 400 Ω, RL = 360 Ω, XL = 200 Ω, ZL R1 = 100 Ω, X2 = 100 Ω, ωM = 80 Ω, 245 ZS = 184 + j0 Ω, VS = ∠0o V 2 M = k L1L 2 M = 0.4 ( 0.125 )( 0.5 ) = 0.1 ⇒ ω = 800 rad/s Fig.33-2 Circuit for Example 33-1 The reflected impedance: Substituting equation (33-4) in equation (33-15) we get Z 22 = R 2 + jωL 2 + ZL = (RL +R2 ) + j ( X2 + XL ) = 400 + j300 Z22 = 400 + j300 = 500Ω and Zr = ω2M2 Z 22 2 ⎡⎣(RL +R2 ) - j ( ωL 2 + XL ) ⎤⎦ -8- EE 205 Dr. A. Zidouri therefore 802 Zr = 400 - j300] =1.62 × 4 - j1.62 ×3 = 10.24 - j7.68 Ω 2 [ 500 802 64 ∗ = = 0.0256 Z The scaling factor by which 22 is reflected is 5002 2500 The primary current I1 is given by equation 33-9 as: Z 22 I1 = VS Z11Z 22 + ω2M2 where Z11 = R1 + jωL1 + ZS = 184 +100 + j400 = 284 + j400Ω and Z22 = 400 + j300 400 + j300 245 o ∠0 Therefore I1 = 2 ( 284 + j400 )( 400 + j300 ) + 80 2 500∠36.87o 245 ∠0o = o 245280∠91.53 + 6400 2 0.5 ∠ - 53.13o A 2 o In the time domain, i1 = 0.5 cos 800t - 53.13 A = ( ) -9- EE 205 Dr. A. Zidouri The secondary current I2 is given by equation 33-10 as: jωM I2 = I1 Z 22 where Z22 = 400 + j300 and ωM = 80 80∠90o 0.5 ∠ - 53.13o Therefore I2 = o 500∠36.87 2 80 0.08 I2 = = A 500 2 2 In the time domain, i2 = 0.08 cos 800t A - 10 - EE 205 Dr. A. Zidouri Self Test: For the following circuit of Fig.33-3 200 500+j100 a 100 j1200 c I1 I2 300 V1 j3600 j1600 V2 b 800-j2500 What is the self-impedance of the primary? a) 500+j100 b) 700+j3700 c) 200+j3600 d) 1000+j4400 Correct answer b What is the self-impedance of the secondary? a) 900-j3700 b) 800-j2500 c) 900-j900 d) 100+j1600 Correct answer c What is the impedance reflected into the primary? a) 800-j2500 b) 100+j1600 c) 1000+j4400 d) 800+j800 Correct answer d d Fig.33-3 for SelfTest What is the impedance seen looking into the primary terminals? a) 700+j4800 b) 200+j3600 c) 1000+j4400 d) 800+j800 Correct answer c - 11 -